If $x_1, x_2,…,x_{10}$ are such that $\sum_{i=1}^{10} \sin^2(x_i) = 1$, prove that $3 \sum_{i=1}^{10} \sin(x_i) \leq \sum_{i=1}^{10} \cos(x_i)$ [duplicate]

Take $x_1, x_2,...,x_{10}$ such that $\sum_{i=1}^{10} \sin^2(x_i) = 1$ with $x_1, x_2,...,x_{10}$ on $\left[0,\frac{\pi}{2}\right]$, prove that $3 \sum_{i=1}^{10} \sin(x_i) \leq \sum_{i=1}^{10} \cos(x_i)$.

I tried to do this: $$\sum_{i=1}^{10} \sin^2(x_i) = \frac{ n-\sum_{i=1}^{10} \cos(2 x_i)}{2} = 1 \rightarrow \sum_{i=1}^{10} \cos(2x_i) = 8$$ So i said $\sum_{i=1}^{10} \cos(2x_i) = A$ and $\sum_{i=1}^{10} \sin(2x_i) = B$. Therefore: $$A + Bi = \sum_{i=1}^{10} \operatorname{cis}(2x_i) = \sum_{i=1}^{10} \operatorname{cis}^2(x_i)$$ But I make no idea about how I will get the inequality. Anybody can help me?

marked as duplicate by Jyrki LahtonenJul 19 '15 at 22:21

• Is "sen" intended to be the sine function, $\sin$? – 6005 Jul 19 '15 at 16:19
• There is an error in your first calculation: $$\sum_{i=1}^n \sin^2(x_i) = \frac{n}{2} \left(1-\sum_{i=1}^n \cos(2 x_i)\right)$$ should be $$\sum_{i=1}^n \sin^2(x_i) = \frac{1}{2} \left(n-\sum_{i=1}^n \cos(2 x_i)\right)$$ – 6005 Jul 19 '15 at 17:35